Suppose $V$ is a vector bundle with structure group $SO(3)$, and suppose that it can be lifted to a $\text{Spin}(3) = SU(2)$ bundle (i.e. $w_2(V) = 0$). Let us call the lifted bundle $E$. Then it is stated on page 42 in The Geometry of Four-Manifolds by Donaldson and Kronheimer that we have the relation $p_1(V) = -4c_2(E)$. My question is:

How does one show this?

More generally, how does one compute the effect of going over to a lifted bundle on the characteristic classes? Generally one has $p_1(E) = c_1^2(E) - 2c_2(E)$. In our case the first term drops out, so that the claim in the book can also be written as $p_1(V) = 2p_1(E)$. This factor $2$ undoubtedly somehow comes from the covering homomorphism $SU(2) \rightarrow SO(3)$ which is 2 : 1, but how?

Probably related: on the same page he defines at the bottom for the so-called instanton number $\kappa = \frac1{8\pi^2}\int_M\text{Tr}(F^2)$, and then claims that this is $\kappa = c_2$ for $SU(r)$ bundles $E$ and $\kappa = -\frac14p_1$ for $SO(r)$ bundles $V$. There again is that factor 4; again, from the formula $p_1(E) = c_1^2(E) - 2c_2(E) = \left[\frac{-1}{4\pi^2}\text{Tr}(F^2)\right]$ one would expect this to be 2. I can see that this factor is chosen so that if one lifts the bundle that $\kappa$ does not change, but on the other hand, not every bundle is liftable, and for bundles which have both Chern classes and Pontryagin classes (such as complex $SU(r)$ bundles), I would expect that one would want the two formulae to give the same answer. As it is, they don't.

(I guess the real problem is I haven't managed to find any readable sources on $SO(r)$-bundles in the context of gauge theories. For special unitary groups there are sources in abundance, but for special orthogonal they are a lot harder to find.)

Johannes answered the question, but one comment that might help if you want a Chern-Weil answer is that $Tr$ means 2 different things. More explicitly, $Tr(F^2)$ is taken in 2x2 matrices in the $su(2)$ case and 3x3 matrices in the $so(3)$ case and the map on lie algebras $c:su(2)\to so(3)$ obtained by differentiating the covering map doesn't preserve this trace. If I recall correctly, for $A\in su(2)$, $Tr_3(c(A)^2)= -2Tr_2(A^2)$, giving your missing factor of 2.
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PaulMar 24 '11 at 13:40

1 Answer
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Let me rephrase your question so that I understand it: let $P \to X$ be an $SU(2)$-principal bundle. Then we get a $2$-dimensional complex vector bundle $E \to X$ by $P \times_{SU(2)} C^2$ (with the defining representation). And we get a $3$-dimensional real vector bundle $V:= P \times_{SU(2)} R^3$ (with the adjoint representation).
You (or Donaldson-Kronheimer) claim that $p_1(V)=-4c_2(E)$.

It is enough to consider the universal case $X=BSU(2)$. The map $BU(1) \to BSU(2)$ induced by the inclusion of the maximal torus is injective in cohomology, so it suffices to check your identity on $BU(1)$ (splitting principle). The pullback of $E$ to $BU(1)$ is $L \oplus L^{\ast}$, so its $c_2$ is a generator $\pm u$ of $H^4 (BU(1))$. The pullback of $V$ to $BU(1)$ is $L^2 \oplus \mathbb{R}$ (here the fact that $SU(2) \to SO(3)$ has degree comes in), hence the total Chern class of $V \otimes \mathbb{C}$ is $(1+2u)(1-2u)$, from which you can read off the Pontrjagin class. For the correct sign, you need a bit more care, though.

EDIT: The inclusion of the maximal torus is the group homomorphism $U(1) \to SU(2)$; $z \mapsto diag (z,z^{-1})$. Thus the defining rep. of $SU(2)$ restricts to a sum of the defining rep. of $U(1)$ and its dual, giving the pullback of $E$.

Consider the standard basis the Lie algebra $\mathfrak{su}(2)$ (http://en.wikipedia.org/wiki/Special_unitary_group#SU.282.29) and compute the action of $diag (z,z^{-1})$ via the adjoint rep in this basis. The result is that the adjoint rep. of $SU(2)$ restricts to a sum of the tensor square of the defining rep of $U(1)$ with the trivial real one-dim. rep. (this computation is basic in representation theory, it is the computation of the roots of $SU(2)$).

Although I'm not really familiar with universal bundles, this seems very helpful. One question (or two, really): why are those pullbacks $L\oplus L^*$ respectively $L^2\oplus\mathbb{R}$? In the first case, applying the splitting principle gives something like $L\oplus K$, so why is $K = L^*$?
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miramoMar 24 '11 at 16:10